| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Easy -1.2 This is a straightforward application of standard formulas for mean and standard deviation from a frequency table, plus a basic probability calculation using combinations. Both parts require only routine recall and calculation with no problem-solving insight needed. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Prize money | \(\pounds 0\) | \(\pounds 10\) | \(\pounds 100\) | \(\pounds 5000\) |
| Frequency | 9929 | 50 | 20 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| P(Two £10 or two £100) \(= \frac{50}{10000} \times \frac{49}{9999} + \frac{20}{10000} \times \frac{19}{9999}\) | M1 | For either correct product seen (ignore any multipliers) |
| M1 | Sum of both correct products (ignore any multipliers) | |
| \(= 0.0000245 + 0.0000038 = 0.000028(3)\) | A1 CAO | As opposite with no rounding |
| After M0,M0: \(\frac{50}{10000} \times \frac{50}{10000} + \frac{20}{10000} \times \frac{20}{10000}\) | SC1 | SC1 case #1 |
| \(\frac{50}{10000} \times \frac{49}{10000} + \frac{20}{10000} \times \frac{19}{10000}\) | SC1 | SC1 case #2 CARE answer is also \(2.83 \times 10^{-5}\) |
| TOTAL | 7 |
## Question 1 (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| P(Two £10 or two £100) $= \frac{50}{10000} \times \frac{49}{9999} + \frac{20}{10000} \times \frac{19}{9999}$ | M1 | For either correct product seen (ignore any multipliers) |
| M1 | Sum of both correct products (ignore any multipliers) |
| $= 0.0000245 + 0.0000038 = 0.000028(3)$ | A1 CAO | As opposite with no rounding |
| After M0,M0: $\frac{50}{10000} \times \frac{50}{10000} + \frac{20}{10000} \times \frac{20}{10000}$ | SC1 | SC1 case #1 |
| $\frac{50}{10000} \times \frac{49}{10000} + \frac{20}{10000} \times \frac{19}{10000}$ | SC1 | SC1 case #2 CARE answer is also $2.83 \times 10^{-5}$ |
| **TOTAL** | **7** | |
---1 A supermarket chain buys a batch of 10000 scratchcard draw tickets for sale in its stores. 50 of these tickets have a $\pounds 10$ prize, 20 of them have a $\pounds 100$ prize, one of them has a $\pounds 5000$ prize and all of the rest have no prize. This information is summarised in the frequency table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Prize money & $\pounds 0$ & $\pounds 10$ & $\pounds 100$ & $\pounds 5000$ \\
\hline
Frequency & 9929 & 50 & 20 & 1 \\
\hline
\end{tabular}
\end{center}
(i) Find the mean and standard deviation of the prize money per ticket.\\
(ii) I buy two of these tickets at random. Find the probability that I win either two $\pounds 10$ prizes or two $\pounds 100$ prizes.
\hfill \mbox{\textit{OCR MEI S1 2009 Q1 [7]}}